Question

A container has milk and water in the ratio 4: 1. The volume of the contents is increased by 40% by adding water to it. From this solution, 5 L are drawn out and replaced with 5 L of milk. If the solution now has five parts milk and three parts water, find the initial volume of the container.​

Answer 1

Answer:

Step-by-step explanation:

Answer is 29.16 litres.

Container contains 40 litres, initially 4 litre milk taken out from that. After that remaining quantity of milk is   40–4=36 litres

Now the container is filled with water . Total quantity is 40 litres out of 40 only 36 litres is milk. If another 4 litres is taken out from the container it is not fully milk. So we have to find quantity if milk   36/40*4=3.6 litres

In 4 litres only 3.6 is milk, now we deduct this also from the total milk quantity  

36–3.6= 32.4litres

Once again the process is repeated  

32.4/40*4=3.24litres

Now deduct this from 32.4litres  

32.4–3.24=29.16litres.

Container contains 29.16litres of milk.

Answer 2

$\Large\text{\underline{\underline{Explanation}}}$

$\Large\text{[Step 1]}$

Let the initial volume of the container be $\text{ x liters}$. It contains milk and water in the ratio of $\large4:1$.

$\text{ \cdots\longrightarrow\dfrac{4x}{5} liters milk and \dfrac{x}{5} liters water.}$

Let's add water to it. Water to be added is $\large40\%$, or $\large\dfrac{2}{5}$ of the initial volume, $\largex$.

So, now in the container is a mixture of

$\text{ \cdots\longrightarrow\dfrac{4x}{5} liters milk and \dfrac{3x}{5} liters water.}$

So, now the ratio of milk and water is $\large4:3$.

$\Large\text{[Step 2]}$

In the mixture of $\large5$ liters taken out is -

$\text{ \cdots\longrightarrow\text{(Milk)}:\text{(Water)}=4:3=\dfrac{20}{7}:\dfrac{15}{7} .}$

So, in the container is a mixture of

$\text{ \cdots\longrightarrow(\dfrac{4x}{5}-\dfrac{20}{7}) liters milk and (\dfrac{3x}{5}-\dfrac{15}{7}) liters water.}$

Now, by replacing the mixture with $\large5$ liters of milk, -

$\text{ \cdots\longrightarrow(\dfrac{4x}{5}+\dfrac{15}{7}) liters milk and (\dfrac{3x}{5}-\dfrac{15}{7}) liters water.}$

$\Large\text{[Step 3]}$

Now we are in the final step.

The mixture now has a $\large5:3$ ratio of milk to water, so let's build an equation.

$\cdots\longrightarrow(\dfrac{4x}{5}+\dfrac{15}{7}):(\dfrac{3x}{5}-\dfrac{15}{7})=5:3$

$\cdots\longrightarrow5\times(\dfrac{3x}{5}-\dfrac{15}{7})=3\times(\dfrac{4x}{5}+\dfrac{15}{7})$

Let's multiply by 35.

$\cdots\longrightarrow5\times(21x-75)=3\times(28x+75)$

$\cdots\longrightarrow105x-375=84x+225$

$\cdots\longrightarrow21x=600$

$\cdots\longrightarrow x=\dfrac{600}{21}$

$\text{ \cdots\longrightarrow \boxed{x=\dfrac{200}{7}\text{ liters}} }$

$\Large\text{[Final answer]}$

Hence, the initial volume of the container is $\large\text{ \dfrac{200}{7} liters}$.